2. Say $3x\equiv 24 (\bmod 60)$. Note that $\gcd(3,60)=3$ and $3|24$, so the congruence is solvable and has exactly $3$ different solutions (up to congruence). If you can find one solution $x_0$, the others are $x_0 + \frac{60}{3}$ and $x_0 + \frac{60}{3}\cdot 2$. So it just remains to find $x_0$. So for example, $x_0=8$ works. Now you can find everything else.